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lina2011 [118]
3 years ago
9

Which descriptor applies to the circuit? A- on B- closed C- incomplete D- open

Physics
2 answers:
gayaneshka [121]3 years ago
8 0
Answer is D- open.hope you get it
HACTEHA [7]3 years ago
3 0
Open circuit is the answer
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PLEASE HELP WILL GIVE BRAINLIEST!!!
Elan Coil [88]

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

\frac{1}{2}kx^2 = mgh

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

g=9.8 m/s^2 is the acceleration due to gravity

h = 3.25 m is the maximum height of the dart

Solving for k, we find:

k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

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6 0
2 years ago
The moment of water from ocean through the atmosphere and back
melisa1 [442]

The water cycle is all about storing water and moving water on, in, and above the Earth. Although the atmosphere may not be a great storehouse of water, it is the superhighway used to move water around the globe. Evaporation and transpiration change liquid water into vapor, which ascends into the atmosphere due to rising air currents. Cooler temperatures aloft allow the vapor to condense into clouds and strong winds move the clouds around the world until the water falls as precipitation to replenish the earthbound parts of the water cycle. About 90 percent of water in the atmosphere is produced by evaporation from water bodies, while the other 10 percent comes from transpiration from plants.

There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

6 0
3 years ago
Read 2 more answers
What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt
ipn [44]
Hope this helps you!

5 0
3 years ago
Read 2 more answers
Two charges are sitting 1.5 m apart with a force of 3 N between them. They are now moved farther apart to 2.25 m and one of the
ANTONII [103]

Answer: 2.37N

Explanation:

According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,

F = kq1q2/r²

For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have

3 = kq1q2/1.5²

3 = kq1q2/2.25

Kq1q2= 6.75... (1)

If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes

F2 = k(4q1)q2/2.25² (q1 has been increased by factor of 4)

k(4q1)q2 = 5.06F2 ... (2)

Dividing 2 by 1 we have

k(4q1)q2/kq1q2 = 5.06F2/3

4 = 5.06F2/3

5.06F2 = 12

F2= 12/5.06

F2 = 2.37N

Therefore the magnitude of the new force between the two charges is 2.37N

5 0
3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is four times as massive as planet 1 and orbits
Keith_Richards [23]
<span>Answer: F=GMm/r^2 where G is the newtonian grav cst, M the mass of the star, m the mass of the planet, and r the planet-star distance F1=GMm1/r1^2 F2=GMm2/r^2 F2/F1 = [GMm2/r2^2]/[GMm1/r1^2] F2/F1=(m2/m1)(r1/r2)^2 = (2)(1/2)^2 = 1/2 because m2/m1=2 and r2/r1=2</span>
8 0
3 years ago
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